Question

помогите ,пожалуйста,номер 11

Answer 1

$sin2x+sqrt{3}sinx = 0 \ 2cos(x)sin(x)+sqrt{3}sinx = 0\ sinx(2cos(x)+sqrt{3})=0 \ sinx = 0 ; x = pi k , k in Z \ cosx = -frac{sqrt{3}}{2}; x = бfrac{5pi}{6}+2pi k$

Answer 2

спасибо большое !

Answer 3

$sin2x+sqrt3sinx=0\\2sinxcdot cosx+sqrt3sinx=0\\sinx(2cosx+sqrt3)=0\\a); ; sinx=0; ,; ; x=pi n,; nin Z\\b); ; 2cosx+sqrt3=0; ,; ; cox=-frac{sqrt3}{2}\\x=pm arccos(-frac{sqrt3}{2})+2pi k=pm (pi -arccosfrac{sqrt3}{2})+2pi k=\\=pm (pi -frac{pi}{6})+2pi k=pm frac{5pi }{6}+2pi k; ,; kin Z\\c); ; xin [-frac{5pi}{2};frac{7pi}{2}]; ,; ; x=0,pi ,; 2pi; ,3pi,; frac{5pi}{6}; ,; frac{7pi}{6}; ,; frac{17pi}{6}; ,; frac{19pi}{6}.$